# Fibonacci Number Proof

1. Nov 8, 2010

### ryou00730

1. The problem statement, all variables and given/known data
The Fibonacci numbers are defined recusively by:
F(0) = 0, F(1) = 1, for n > 1, F(n) = F(n − 1) + F(n − 2).
Use strong induction to show that F(n) < 2^n for all n.

2. Relevant equations
n/a

3. The attempt at a solution
I use my base case as F(2) = F(1) + F(0) = 1 which is less than 2^n = 2^2 = 4. After that I am not sure where to go with strong induction...

2. Nov 8, 2010

### micromass

Staff Emeritus
The induction hypthesis gives us

$$F(n)=F(n-1)+F(n-2)\leq 2^{n-1}+2^{n-2}$$

Can you now prove that this is smaller then $$2^n$$?

3. Nov 8, 2010

### ryou00730

well 2^n-1 + 2^n-2 is equal to 2^n (2^-1 + 2^-2) = 3/4(2^n) but I don't see the logic behind equating the problem to less than or equal to 2^n-1 + 2^n-2

4. Nov 8, 2010

### micromass

Staff Emeritus
If you want to apply induction? Then what is the induction hypothesis?

5. Nov 8, 2010

### ryou00730

Let k>2, for all integers i, with 2<=i<k, F(i)<2^i would be my induction hypothesis I think?

6. Nov 8, 2010

### micromass

Staff Emeritus
Yes, that is correct. So in particular, we have

$$F(n-1)\leq 2^{n-1}~\text{and}~F(n-2)\leq 2^{n-2}$$

So

$$F(n-1)+F(n-2)\leq 2^{n-1}+2^{n-2}$$

7. Nov 8, 2010

### ryou00730

woops, I meant n>2, so just replace all my k's with n's and so if I know that F(i)<2^i, then F(i-1) + F(i-2) < 2^i-1 + 2^i-2 which is F(i) < 3/4 2^i, so therefore F(i) must also be less than 2^i if its less than 0.75 of it, so the statement is true for all F(n)

8. Nov 8, 2010

9. Nov 8, 2010

### micromass

Staff Emeritus
Yeah, that looks fine

10. Nov 8, 2010

### ryou00730

thank you :), so in general with strong induction, you prove a base, then your induction hypothesis is that it works for all numbers between your base up until some value n, and you have to prove using this, that it also works for the n? thank you for all your help!

11. Nov 8, 2010

### micromass

Staff Emeritus
Yep, that is the idea behind strong induction.

Good luck with your next problems!

12. Nov 8, 2010

thank you:)